Let f(x) = `{{:( x^(3) + x^(2) - 10 x ,, -1 le x lt 0) , (sin x ,, 0 le x lt x//2) , (1 + cos x ,, pi //2 le x le x ):}` then f(x) has

Options

A. local maximum at ` x = pi//2`
B. local minima at `x = pi//2`
C. absolute minima at x = `0 , pi`
D. absolute maxima at `x = pi //2`

Answer

To solve the problem, we will analyze the piecewise function \( f(x) \) defined as follows:

\[ f(x) = \begin{cases} x^3 + x^2 - 10x & \text{for } -1 \leq x < 0 \\ \sin x & \text{for } 0 \leq x < \frac{\pi}{2} \\ 1 + \cos x & \text{for } \frac{\pi}{2} \leq x \leq \pi \end{cases} \]

### Step 1: Find the derivative \( f'(x) \) for each piece of the function.

1. For \( -1 \leq x < 0 \):
\[ f'(x) = 3x^2 + 2x - 10 \]

2. For \( 0 \leq x < \frac{\pi}{2} \):
\[ f'(x) = \cos x \]

3. For \( \frac{\pi}{2} \leq x \leq \pi \):
\[ f'(x) = -\sin x \]

### Step 2: Set the derivatives to zero to find critical points.

1. For \( -1 \leq x < 0 \):
\[ 3x^2 + 2x - 10 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-10)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 120}}{6} = \frac{-2 \pm \sqrt{124}}{6} = \frac{-2 \pm 2\sqrt{31}}{6} = \frac{-1 \pm \sqrt{31}}{3} \]
We need to check which of these values lie in the interval \([-1, 0)\).

2. For \( 0 \leq x < \frac{\pi}{2} \):
\[ \cos x = 0 \implies x = \frac{\pi}{2} \]

3. For \( \frac{\pi}{2} \leq x \leq \pi \):
\[ -\sin x = 0 \implies x = \pi \]

### Step 3: Evaluate the critical points and endpoints.

1. For \( x = -1 \):
\[ f(-1) = (-1)^3 + (-1)^2 - 10(-1) = -1 + 1 + 10 = 10 \]

2. For \( x = 0 \):
\[ f(0) = \sin(0) = 0 \]

3. For \( x = \frac{\pi}{2} \):
\[ f\left(\frac{\pi}{2}\right) = 1 + \cos\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \]

4. For \( x = \pi \):
\[ f(\pi) = 1 + \cos(\pi) = 1 - 1 = 0 \]

### Step 4: Compare the function values to find absolute maxima and minima.

- \( f(-1) = 10 \)
- \( f(0) = 0 \)
- \( f\left(\frac{\pi}{2}\right) = 1 \)
- \( f(\pi) = 0 \)

### Conclusion:

- The **absolute maximum** occurs at \( x = -1 \) with \( f(-1) = 10 \).
- The **absolute minimum** occurs at \( x = 0 \) and \( x = \pi \) with \( f(0) = 0 \) and \( f(\pi) = 0 \).

### Final Answer:
- **f(x) has an absolute maximum at \( x = -1 \) and absolute minima at \( x = 0 \) and \( x = \pi \).**